LibreOJ 3298.「BJOI2020」封印

明明是唯一一道区分题能不能认真点啊喂(
怎么出得这么 naive 啊(

首先用任何一种后缀数据结构对 \(1 \le i \le |S|\) 求出 \[ l_i = \max\{j \mid S_{i - j + 1\dots i} \in {\rm substrings}(T)\} \]

那么容易发现答案即是 \(\max\{\min(l_i,i - l + 1) \mid l \le i \le r\}\)
讨论一下 \(i - l_i + 1\) 的大小,容易发现其有单调性,可以二分然后 RMQ。
线性的话离线双指针然后 \(O(1)\) RMQ 即可。

代码:

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#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 2e5;
const int LG = 20;
int n,m,q,lg2[N + 5];
char s[N + 5],t[N + 5];
int len[N + 5];
int st[LG + 5][N + 5];
inline int query(int l,int r)
{
int lg = lg2[r - l + 1];
return max(st[lg][l],st[lg][r - (1 << lg) + 1]);
}
int ans;
namespace SAM
{
struct node
{
int ch[2];
int fa,len;
} sam[(N << 1) + 5];
int las = 1,tot = 1;
inline void insert(int x)
{
int cur = las,p = ++tot;
sam[p].len = sam[cur].len + 1;
for(;cur && !sam[cur].ch[x];cur = sam[cur].fa)
sam[cur].ch[x] = p;
if(!cur)
sam[p].fa = 1;
else
{
int q = sam[cur].ch[x];
if(sam[cur].len + 1 == sam[q].len)
sam[p].fa = q;
else
{
int nxt = ++tot;
sam[nxt] = sam[q],sam[nxt].len = sam[cur].len + 1,sam[p].fa = sam[q].fa = nxt;
for(;cur && sam[cur].ch[x] == q;cur = sam[cur].fa)
sam[cur].ch[x] = nxt;
}
}
las = p;
}
inline void build()
{
for(register int i = 1;i <= m;++i)
insert(t[i] - 'a');
for(register int i = 1,x,p = 1,l = 0;i <= n;++i)
{
x = s[i] - 'a';
if(sam[p].ch[x])
p = sam[p].ch[x],++l;
else
{
for(;p && !sam[p].ch[x];p = sam[p].fa);
!p ? (p = 1,l = 0) : (l = sam[p].len + 1,p = sam[p].ch[x]);
}
st[0][i] = len[i] = l;
}
}
}
inline int get(int L,int R)
{
int l = L,r = R,mid,ret = R + 1;
while(l <= r)
mid = l + r >> 1,mid - len[mid] + 1 >= L ? (r = mid - 1,ret = mid) : (l = mid + 1);
return ret;
}
int main()
{
scanf("%s%s%d",s + 1,t + 1,&q),n = strlen(s + 1),m = strlen(t + 1);
for(register int i = 2;i <= n;++i)
lg2[i] = lg2[i >> 1] + 1;
SAM::build();
for(register int i = 1;(1 << i) <= n;++i)
for(register int j = 1;j + (1 << i) - 1 <= n;++j)
st[i][j] = max(st[i - 1][j],st[i - 1][j + (1 << i - 1)]);
for(int l,r,x;q;--q)
{
scanf("%d%d",&l,&r),x = get(l,r),ans = 0;
x > l && (ans = max(ans,x - l));
x <= r && (ans = max(ans,query(x,r)));
printf("%d\n",ans);
}
}